Poisson, Binomial Distributions

The number of claims that an insurance company receives per week is a random variable with the Poisson distribution with parameter λ. The probability that a claim will be accepted as genuine is p, and is independent of other claims.

a) What is the probability that no claim will be accepted over one week?
b) Find the expected number of accepted claims over one week.
c) Let N be the number of accepted claims over one week. Find the probability distribution for N.

2. Relevant equations

Poisson: P(X=x) = λx/x! e-λ

3. The attempt at a solution

a) If x is the number of attempted claims, P(N = 0) = (1 - p)x, I think.

b/c) The distribution for N should be binomial(x, p). Now the expectation value of X is λ. And the expectation value of N is xp. Would this then become λp?

The number of claims that an insurance company receives per week is a random variable with the Poisson distribution with parameter λ. The probability that a claim will be accepted as genuine is p, and is independent of other claims.

a) What is the probability that no claim will be accepted over one week?
b) Find the expected number of accepted claims over one week.
c) Let N be the number of accepted claims over one week. Find the probability distribution for N.

2. Relevant equations

Poisson: P(X=x) = λx/x! e-λ

3. The attempt at a solution

a) If x is the number of attempted claims, P(N = 0) = (1 - p)x, I think.

b/c) The distribution for N should be binomial(x, p). Now the expectation value of X is λ. And the expectation value of N is xp. Would this then become λp?

Yes.

I mean: the expectation is ##\lambda p##. You still need to deal with the issue of the probability values---not just the expected value.